Question

Find the value of the series $x{{\log }_{e}}a+\dfrac{{{x}^{3}}}{3!}{{\left( {{\log }_{e}}a \right)}^{3}}+\dfrac{{{x}^{5}}}{5!}{{\left( {{\log }_{e}}a \right)}^{5}}+...$\[\]
A. $\cosh \left( x{{\log }_{e}}a \right)$\[\]
B. $\coth \left( x{{\log }_{e}}a \right)$\[\]
C $\sinh \left( x{{\log }_{e}}a \right)$\[\]
D. $\tanh \left( x{{\log }_{e}}a \right)$\[\]

Answer 1

Hint: Use the exponential series $ {{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$ and the definition of hyperbolic trigonometric functions $\left( \sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2},\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)$ to test whether Option A and C is correct. Use the Taylor series expansion whether option B and D is correct. \[\]

Complete step-by-step answer:
We know that the base of natural logarithm $e$ is defined as $e=\underset{n\to 0}{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{n} \right)}^{n}}$. We take some real number $x$ and for some $n>1$ we use the expansion of ${{\left( 1+\dfrac{1}{n} \right)}^{nx}}$. So
\[\begin{align}
  & {{\left( 1+\dfrac{1}{n} \right)}^{nx}}=1+nx\times \dfrac{1}{n}+\dfrac{nx\left( nx-1 \right)}{2!}\times \dfrac{1}{{{n}^{2}}}+... \\
 & \Rightarrow {{\left\{ {{\left( 1+\dfrac{1}{n} \right)}^{n}} \right\}}^{x}}=1+x+\dfrac{x\left( x-\dfrac{1}{n} \right)}{2!}+... \\
\end{align}\]
Now let us take the limit $n\to \infty $ above and get,
\[{{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...\]
Which is called the exponential series or the expansion of ${{e}^{x}}$. We find the expansion of ${{e}^{-x}}$ by putting $-x$ in the above expansion. We have,
\[\begin{align}
  & {{e}^{-x}}=1+\dfrac{\left( -x \right)}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{\left( -x \right)}^{3}}}{3!}+... \\
 & \Rightarrow {{e}^{-x}}=1-x+\dfrac{{{x}^{2}}}{2!}-{{x}^{3}}+... \\
\end{align}\]
The hyperbolic functions are analogues of trigonometric functions defined for hyperbola instead of circle. The parametric form of coordinate in two dimensions $\left( \cos t,\sin t \right)$ which form a circle of unit radius. Similarly the parametric point $\left( \cosh t,\sinh t \right)$ forms the right half of the equilateral hyperbola. The definition of basic hyperbolic trigonometric functions in terms of exponential function in the a parametric form is given as
\[ \begin{align}
  & \sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \\
 & \cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \\

\end{align}\]
The given series is
\[x{{\log }_{e}}a+\dfrac{{{x}^{3}}}{3!}{{\left( {{\log }_{e}}a \right)}^{3}}+\dfrac{{{x}^{5}}}{5!}{{\left( {{\log }_{e}}a \right)}^{5}}+...\]
The above can also be expressed as
\[x{{\log }_{e}}a+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{3}}}{3!}+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{5}}}{5!}+...\text{ }....(1)\]
So let us check each option given in the question. We need the exponential expansion of ${{e}^{x{{\log }_{e}}a}},{{e}^{-x{{\log }_{e}}a}},{{e}^{2}}^{x{{\log }_{e}}a}$. So we have,
\[\begin{align}
  & {{e}^{x{{\log }_{e}}a}}=1+\dfrac{x{{\log }_{e}}a}{1!}+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{2}}}{2!}+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{3}}}{3!}+... \\
 & {{e}^{-x{{\log }_{e}}a}}=1-\dfrac{x{{\log }_{e}}a}{1!}+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{2}}}{2!}-\dfrac{{{\left( x{{\log }_{e}}a \right)}^{3}}}{3!}+... \\
\end{align}\]
Checking Option A:\[\]
We see in this option hyperbolic cosine function is given. Using the definition of hyperbolic cosine we get

\[\cosh \left( x{{\log }_{e}}a \right)=\dfrac{{{e}^{x{{\log }_{e}}a}}+{{e}^{x{{\log }_{e}}a}}}{2}=2\left( 1+\dfrac{{{x}^{2}}{{\log }_{e}}a}{2!}+\dfrac{{{x}^{4}}{{\log }_{e}}a}{4!}+... \right)\]
We see that the simplified series is not the same as series (1). So option A is incorrect.
Checking option B.\[\]
We see in this option a hyperbolic cotangent function is given. We cannot expand the cotangent with exponential series. We need Taylor ‘s series expansion of co-tangent function which is defined for some $x$ lies between 0 and $\pi $as $\dfrac{1}{x}+\dfrac{x}{3}-\dfrac{{{x}^{3}}}{45}+..$. . If we shall put $x=x{{\log }_{e}}a$ , it will not be the same as the series(1). So option B is incorrect. \[\]
Checking Option C\[\]
Here we are given the sine hyperbolic function.
\[\sinh \left( x{{\log }_{e}}a \right)=\dfrac{{{e}^{x{{\log }_{e}}a}}-{{e}^{-x{{\log }_{e}}a}}}{2}=x{{\log }_{e}}a+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{3}}}{2!}+\dfrac{{{\left( x{{\log }_{e}}a \right)}^{5}}}{5!}+...\]
The obtained series is the same as series(1) . So the correct option is C.\[\]

Note: We can also check the option D with Taylor’s series expansion of tangent function which defied for some $\left| x \right|<\dfrac{\pi }{2}$ as $x-\dfrac{{{x}^{3}}}{3}+\dfrac{2{{x}^{3}}}{45}+...$ . We can also express but not expand hyperbolic tangent and cotangent function as $\tanh x=\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1},\coth x=\dfrac{{{e}^{2x}}+1}{{{e}^{2x}}-1}$